$\dfrac{d}{dx}[(8+7x)^4]=\,?$ Choose 1 answer: Choose 1 answer: (Choice A) A $4(8+7x)^3$ (Choice B) B $4(7)^3$ (Choice C) C $28(8+7x)^3$ (Choice D) D $28x^3$
Since $(8+7x)^4$ is a composite function, we can use the chain rule. The chain rule says: $\dfrac{d}{dx}\left[w\Bigl(u(x)\Bigr)\right]={w'\Bigl({u(x)}\Bigr)}\cdot{u'(x)}$ To use it, we need to recognize our function as a composite function like $w\big(u(x)\big)$. $\underbrace{(~\overbrace{8+7x}^{\text{inner}}~)^4}_{\text{outer}}$ So if $(8+7x)^4=w(u(x))$, then: $\begin{aligned} {u(x)}&={8+7x} &&\text{inner function} \\\\ w(x)&=x^4&&\text{outer function} \end{aligned}$ Before applying the chain rule, let's find the derivatives of the inner and outer functions (work not shown, but hopefully these derivatives are familiar by now). $\begin{aligned} {u'(x)}&={7} \\\\ {w'(x)}&={4x^3} \end{aligned}$ Now let's apply the chain rule: $\begin{aligned} \dfrac{d}{dx}[(8+7x)^4]&=\dfrac{d}{dx}\left[w\Bigl(u(x)\Bigr)\right] \\\\ &={w'\Bigl({u(x)}\Bigr)}\cdot{u'(x)} \\\\ &={4({8+7x})^3} \cdot {7} \\\\ &=28(8+7x)^3 \end{aligned}$